find the probability that a point chose a t random on the part of the number line shown will lie between points B and C. (anwser should be in a fraction or decimal)

The amount made in commission is $67.5.

The amount made in sales by the store is $682.5.

The percentage is calculated by dividing the required value by the total value and multiplying by 100.

Example:

Required percentage value = a

total value = b

Percentage = a/b x 100

Example:

50% = 50/100 = 1/2

25% = 25/100 = 1/4

20% = 20/100 = 1/5

10% = 10/100 = 1/10

We have,

Amount sold last week = $750.

Commission = 9%

The amount of commission.

= 9/100 x 750

= $67.5

The sales made last week by the store.

= 750 - 67.5

= $682.5

Thus,

$67.5 was made in commission.

$682.5 was made from sales.

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We would need to multiply the first equation by -4 to eliminate the y variable when solving the system of equations by elimination.

The first equation is: 3x + 4y = 5

To eliminate the y variable when solving the system of equations by elimination, we need to multiply the first equation by -4. This is because when two equations are multiplied by the same number, any terms that have the same variable will be eliminated when the equations are added together.

Formula:

3x + 4y = 5

-4(3x + 4y = 5)

3x + 4y = 5

-12x - 16y = -20

Thus, we would need to multiply the first equation by -4 to eliminate the y variable when solving the system of equations by elimination.

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The estimated perimeter of an ellipse if the major axis has a length of 15 ft and the minor axis has a length of 7.5 ft id found to be 37.3 feet .

Perimeter is calculated as

= 2*pi*r*r

=  2 *3.14 * sqrt (15/2² +7.5/2²)/2

= 6.28 x sqrt (56.25+14.0625/2 6.28) *sqrt (35.15625) 6.28 * 5.929270613

= 37.235 feat

An ellipse is the locus of all the points on a plane whose distances from two fixed points in the plane are always same. The fixed points, which are encompassed by the curve, are known as foci , singular of focus.

The constant ratio is the eccentricity of the ellipse and the fixed line is directrix. Eccentricity is an element of  ellipse which denotes elongation and is symbolized by the letter 'e'.

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The factor of expression 3x² 12xy is 3x². To find the factor, we need to divide 12xy by 3x². We start by dividing the coefficients, 12 divided by 3 is 4. Then we divide the x terms, x divided by x is 1. Finally, we divide the y terms, y divided by y is 1. Therefore, the factor of 3x² 12xy is 3x².

The factor of  expression 3x² 12xy is 3x². To find the factor, we need to divide 12xy by 3x². We start by dividing the coefficients, 12 divided by 3 is 4. Then we divide the x terms, x divided by x is 1. This means that the x part of the factor is 3x. Next, we divide the y terms, y divided by y is 1. This means that the y part of the factor is y. When we combine the two parts, 3x and y, we get the factor of 3x². Therefore, the factor of 3x² 12xy is 3x². This means that 3x² is a factor of 12xy, which can be seen by multiplying 3x² by 4y, which results in 12xy. This shows that 3x² is a factor of 12xy.

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The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]

[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]

Eigenspace corresponding to lambda=5,1,4

The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]

for lambda=5

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]

Reducing the matrix A-5I by elementary row operations

[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]

[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]

[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]

The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix

[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis

[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]

we reduce the matrix A-I by elementary row operations as follows.

[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]

[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]

Thus, the solutions x of (A-I) X=0 satisfy

[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]

[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]

The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]

case:3

lambda=4

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]

we reduce the matrix A-4I by elementary row operations as follows.

[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]

[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]

Thus, the solutions x of (A-4IX)=0 satisfy

[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]

[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]

The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]

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The 3 types of terms in quadratic equation:

1) quadratic term,

2) linear term,

3) constant term.

We know that the second degree algebraic equation in x is a quadratic equation.

The standard form of quadrtic equation is ax^2 + bx + c = 0, where a, b, c are integers.

As thise is a quadratic equation, the value of a can not be zero. a ≠ 0

The term ax^2 is called the quadratic term.

From this term, the name given to the equation(quadrtic equation)

The term bx is called the linear term.

And the term c is called the constant term.

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Answer: slope=2

Step-by-step explanation:

using y=mx+b, we know the equation of the line is y=2x. The slope is the m term so we know that the slope is 2.

On solving the provided question, we can say that the required probability is = 13/204.

Probability theory, a subfield of mathematics, gauges the likelihood of an occurrence or a claim being true. An event's probability is a number between 0 and 1, where approximately 0 indicates how unlikely the event is to occur and 1 indicates certainty. A probability is a numerical representation of the likelihood or likelihood that a particular event will occur. Alternative ways to express probabilities are as percentages from 0% to 100% or from 0 to 1. the percentage of occurrences in a complete set of equally likely possibilities that result in a certain occurrence compared to the total number of outcomes.

probability of 1st card =  1/4

since the card is not replaced

total number remaining cards = 51

second card 13/51

the required probability is = 13/204

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Answer:

The total cost of the science class trip to the zoo is $1200

Step-by-step explanation:

The cost of the science class trip to the zoo is $1200, and the cost of Mr. Sexton's class trip to the state park is $1000.

Answer:

n = - 1 , n = 3

Step-by-step explanation:

n² - 2n - 3 = 0 ( add 3 to both sides )

n² - 2n = 3

to complete the square

add ( half the coefficient of the x- term)² to bpth sides

n² + 2(- 1)n + 1 = 3 + 1

(n - 1)² = 4 ( take square root of both sides )

n - 1 = ± [tex]\sqrt{4}[/tex] = ± 2 ( add 1 to both sides )

n = 1 ± 2

Then

n = 1 - 2 = - 1

n = 1 + 2 = 3

Answer:

7. .44

8. .009

Step-by-step explanation:

7. 44 ÷ 100 = .44

8. 9 ÷ 1000 = .009

By drawing a number line and then marking it as 0 and 1, and by dividing the distance in 3 equal parts we can graph 1/3 on the number line, where each part will be equal to the 1/3.

A number line is a diagram of a graded straight line used to represent real numbers in introductory mathematics. It is assumed that every point on a number line corresponds to a real number, and that every real number corresponds to a point.

A horizontal line with evenly spaced numerical increments is referred to as a number line. How the number on the line can be answered depends on the numbers present.

For drawing 1/3 on a number line, we will follow the following steps:

We will first draw a line and mark 0 and 1 on it.

In between 0 and 1, we will divide the total distance into 3 equal parts, where one part will represent (1/3)th portion.

Hence. by marking there we can locate 1/3 on the number line.

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The correct question may be like:

How do you plot 1/3 on a graph.

The irrational numbers to children can be explained as the numbers  that can be written as a nonrepeating or nonterminating decimal .

What are Irrational Numbers ?

The real numbers that can be represented as a nonrepeating or a nonterminating decimal but not as a fraction, and the decimal that goes on forever without repeating.

For Example : [tex]\sqrt{2} , \sqrt{5} , \sqrt{7}[/tex] are few example of irrational numbers .

In simpler words : the irrational number is a number that is not rational. which means It is a number which cannot be written as a ratio of two integers or cannot be written as fraction.

and If a fraction has a 0 in the denominator , it is an irrational number .

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The properties of the rational exponents are given and a rational equation is of the form b = aˣ

When you raise a quotient to a power you raise both the numerator and the denominator to the power. When you raise a number to a zero power you'll always get 1. Negative exponents are the reciprocals of the positive exponents.

The different Laws of exponents are:

mᵃ×mᵇ = mᵃ⁺ᵇ

mᵃ / mᵇ = mᵃ⁻ᵇ

( mᵃ )ᵇ = mᵃᵇ

mᵃ / nᵃ = ( m / n )ᵃ

m⁰ = 1

m⁻ᵃ = ( 1 / mᵃ )

Given data ,

Let the rational exponent equation be A

Now , the properties of the exponent equations are

mᵃ×mᵇ = mᵃ⁺ᵇ

The powers of the exponents are added together

mᵃ / mᵇ = mᵃ⁻ᵇ

The powers of the exponents are subtracted together

( mᵃ )ᵇ = mᵃᵇ

The powers of the exponents are multiplied together

mᵃ / nᵃ = ( m / n )ᵃ

m⁰ = 1

Any number raised to the power of 0 is 1

m⁻ᵃ = ( 1 / mᵃ )

Hence , the exponents are solved

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Answer:

6 1/4 pounds

Step-by-step explanation:

There are 7 1/2 pounds of potatoes, which is equivalent to 7 1/2 = 7.516=120

120 ounces of potatoes.

If 1/6 of the potatoes are rotten, then 1/6*120 = 20

20 ounces of potatoes are rotten.

Thus, the weight of the good potatoes is 120-20 = 100

100 ounces.

Converting this back to pounds, we get

100/16 = 6 1/4

6 1/4 pounds.

Answer:

so i will be honest I've done this before and it may be 64 or 120

Step-by-step explanation:

It may be one of these answers I'm doing as much as I can to help

Of the part, the whole, and the percent, a. the exact number of students that are athletes is unknown to Ms. Morgan, which is 36, and b. the total strength of the students is unknown to Mr. Gregory which is 180.

Here we need to know 3 things- the total no. of students each teacher has, the percentage of athletes, and no. of students that are athletes.

a)

As we can see in the case of Ms. Morgan, we know the total strength of the students and the percentage of the students that are Athletes. Hence what we don't know here is the exact number of students that are athletes.

Hence we need to find 30% of 120

= 30/100 X 120

= 36 students.

b)

Here, in Mr. Gregory's case, we know the percentage and the actual number of students that are athletes. Hence, here we don't know the total strength of the students Mr. Gregory has.

Total strength = no. of athletes X 100/percent of athletes

= 27 X 100/15

= 180 students.

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Complete Question

Ms. Morgan is a science teacher. She found that 30% of her 120 students are athletes. Mr. Gregory is a math teacher. He found that 27, or 15%, of his students, are athletes. a. Of the part, the whole, and the percent, which information is still unknown to Ms. Morgan? Find this unknown value. b. Of the part, the whole, and the percent, which information is still unknown to Mr. Gregory? Find this unknown value.

The  set of equations correctly representing x are as follows:

(x + 1/2) and (x = 7/2).

A set or group of equations that you solve all at once is referred to as a "system" of equations. The simplest linear system is one that has two equations as well as two variables. Linear equations (those that graph as straight lines) are easier to understand than non-linear ones.

Systems of equations in mathematics are a collection of relationships between different unknown variables that can be stated in terms of algebraic expressions. They are also known as simultaneous equations. Graphing, substitution, as well as elimination by addition, are methods that can be used to find the solutions to a basic system of equations.

Sum means addition (+).

Given that sum of a number, x and 1/2 is 4.

This means the adding  two numbers

x+1/2=4

Now make x the subject of formula,

x=4-1/2

Finding the L.C.M  = 8-1/2

                        x=7/2

                     ∴ x=3 1/2

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According to the following graph, the reflection point of (-3,4) across the line y = 2 is (-3,0).

The term reflection point in math is defined as

Here we need to find the reflection point of (-3, 4) across the line y = 2.

Here first we have t plot the point on the graph, and then we have to plot the line equation y = 2 on the graph,

Here in order to find the reflection of a point along the x-axis, then we have to keep the abscissa constant and see the reflection of ordinate along the x-axis.

Based on these rule, the resulting graph is obtained as follows.

Through the graph we have identified that the resulting reflection point is (-3,0).

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let's hmmm hmmm multiply both sides by the LCD of all denominators, that way we do away with the denominators and see what "x" is

[tex]\cfrac{x+1}{2}+\cfrac{x+2}{3}=3-\cfrac{x+3}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{x+1}{2}+\cfrac{x+2}{3} \right)=12\left( 3-\cfrac{x+3}{4} \right)} \\\\\\ (6x+6)+(4x+8)=36-(3x+9)\implies 10x+14=36-3x-9 \\\\\\ 10x+14=27-3x\implies 10x=13-3x\implies 13x=13 \\\\\\ x=\cfrac{13}{13}\implies x=1 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{1-4x}{10}-\cfrac{2x+1}{2}=\cfrac{5x+1}{5}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{10}}{10\left( \cfrac{1-4x}{10}-\cfrac{2x+1}{2}\right)=10\left( \cfrac{5x+1}{5} \right)} \\\\\\ (1-4x)-(10x+5)=10x+2\implies 1-4x-10x-5=10x+2 \\\\\\ -14x-4=10x+2\implies -4=24x+2\implies -6=24x \\\\\\ \cfrac{-6}{24}=x\implies -\cfrac{1}{4}=x[/tex]

27 2/10

   or

27 1/5

----------------------------------

On solving the provided question, we can say that - here in number line we jump fron 2 to 7 as we have added +5

A number line is a visual representation of real numbers used in introductory mathematics. It is an image of a magnitude line. On the number line, each point represents a real number, and each real number is taken to represent a point. Increments on a number line are separated by equal distances. You can only respond to the numbers on a line in the manner indicated by those numbers. How the number is utilized is determined by the question that goes with it. B: Make a point.

here,

we have +5

so 2+(+5) = 7

we will jump to 7 from 2

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The solution to the polynomial is x = -1, x = 1/2 and x = -1

From the question, we have the following parameters that can be used in our computation:

2x³ - x² - 2x + 1

Expand

2x³ - x² - 2x + 1 = 2x³ - x² - 2x + 1

Factorize the expression

This gives

2x³ - x² - 2x + 1 = x²(2x - 1) - 1(2x - 1)

Factor out 2x - 1

2x³ - x² - 2x + 1 = (x² - 1)(2x - 1)

Express x² - 1 as difference of two squares

2x³ - x² - 2x + 1 = (x - 1)(x + 1)(2x - 1)

So, we have

(x - 1)(x + 1)(2x - 1) = 0

Solve for x

x = 1, x = -1 and x = 1/2

Reorder the solutions

x = -1, x = 1/2 and x = -1

Hence, the solution is x = -1, x = 1/2 and x = -1

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Answer: The Gradient of line A and B are

2 and -1

Step-by-step explanation:

For the given two points A(x1, y1) and B(x2, y2)

The gradient of the line AB is y2-y1/x2-x1

So the gradient of line A is

5-1/2-0

2

the gradient of line B is

5-0/0-5

-1

Angle ADC = 120 degrees.

A certain portion of a circle that is created based on two radius of the same circle and one arc. Area of sector for a circle with radius r is given by π r²Ф/ 360°

Given, two circles of radius 1 unit and 2 unit which have same center D.

We know area of a circle = π r²

Area of larger circle = π 2² = 4π

it is said that the total area of the shaded region that means area of a particular sector is 1/12 of the area of the larger circle.

area of sector, ACD = 1/12 ×4π

as per the question, the ACD sector is located in the smaller circle that has radius 1 unit.

formula for area of sector ACD = π r² ×Ф/360°

where,Фis the central angle of ACD sector

and Ф = ADC

from the above statement, 4π/12 = π r² ADC/360

                        ADC = 1/3×360°

ADC = 120 degrees

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The inequalities that represent the situation are 15x + 10y ≥ 400 and x + y ≤ 35 and the viable solutions are (10, 25), (20, 12), (35, 0) and (20, 20)

From the question, we have the following parameters that can be used in our computation:

These parameters above mean that

15x + 10y = Total earnings

x + y = Number of hours

So, we have

15x + 10y ≥ 400

x + y ≤ 35

The above represent the inequalities of the situation

In (a), we have

15x + 10y ≥ 400

x + y ≤ 35

Next, we test the options

(10, 25)

15 * 10 + 10 * 25 ≥ 400 ⇒ 400 ≥ 400

10 + 25 ≤ 35 ⇒ 35 ≤ 35

True

(10, 20)

15 * 10 + 10 * 20 ≥ 400 ⇒ 350 ≥ 400

10 + 20 ≤ 35 ⇒ 30 ≤ 35

False

(20, 12)

15 * 20 + 10 * 12 ≥ 400 ⇒ 420 ≥ 400

20 + 12 ≤ 35 ⇒ 32 ≤ 35

True

(35, 0)

15 * 35 + 10 * 0 ≥ 400 ⇒ 525 ≥ 400

35 + 0 ≤ 35 ⇒ 35 ≤ 35

True

(20, 20)

15 * 20 + 10 * 20 ≥ 400 ⇒ 500 ≥ 400

20 + 20 ≤ 35 ⇒ 40 ≤ 35

False

Hence, the viable solutions are (10, 25), (20, 12), (35, 0) and (20, 20)

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Answer:

2

Step-by-step explanation:

0.2(x+50)-6=0.4(3x+20) multiply both sides by 10 to clear the decimals

2(x + 50) = 4(3x + 20) distribute across the terms in the parentheses

2x + 100 = 12x + 80 subtract 80, 2x from both sides

20 = 10x divide both sides by 10

x=2

If 1 Swedish krona was worth 0.12 US dollar , and if he took d dollars for the trip then the equation that is used to find the value in Swedish krona (k) of the dollars is option (a) [tex]k = 0.12\times d[/tex] .

The value of 1 Swedish krona is = 0.12 US dollar ,

the number of dollar that Souta took on the trip is = d dollars ;

we have to find the equation that can be used to find the equation in Swedish krona "k" of the dollars ,

According to the question ,

the equation for Swedish krona can be represented as [tex]k = 0.12\times d[/tex] ;

Therefore , the required equation is [tex]k = 0.12\times d[/tex] .

The given question is incomplete , the complete question is

When Souta went to Sweden, 1 Swedish krona was worth about 0.12 US dollar. He took d dollars on the trip. Which equation could be used to find the value in Swedish krona, k, of the dollars ?

(a) k = 0.12d

(b) dk = 0.12

(с) 0.12k = d

(d) k = 1.12d

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The length AC will be 5.83 and the angle ACD will be 34.45°.

The branch of mathematics that sets up a relationship between the sides and the angles of the right-angle triangle are termed trigonometry.

The trigonometric functions, also known as a circular, angle, or goniometric functions in mathematics, are real functions that link the angle of a right-angled triangle to the ratios of its two side lengths.

The length AC will be calculated as,

AC² = 5² + 3²

AC = √ ( 25 + 9 )

AC = √34

AC = 5.83

The angle ACD will be,

tanθ = P / B

θ = tan⁻¹ = ( 4 / 5.83)

θ = 34.45°

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Step 3 is Define the problem Goals.

there are 8 steps, which are as follows,

1- Define the Problem

2- Clarification of the Problem

3- Define the problem Goals

4- Identify main Cause of the Problem

5- Develop a Action Plan

6- Execute that  Action Plan

7- Analyze the Results

8- Continuous Improvements

Problem solving

Problem solving is the method of determining the problem and then clarify the doubts and prepare a action plan and then execute the action plan and find out the desired results and if the results are not appropriate then improve the action plan and your way of doing until you get the desired results.

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